GMATGuruNY wrote:

edge wrote:

GMATGuruNY wrote:...Total arrangements with AB and CD together means we're arranging the 3 elements AB, CD, and E = 3! = 6. Now we multiply by 4 because AB can be reversed, CD can be reversed, and both AB and CD can be reversed: 4*6 = 24...

Shouldn't we multiply by 8 because of 3 reversals?

The number of ways to arrange the 3 elements AB, CD and E = 3! = 6.
There are 4 variations in which AB and CD sit as couples: AB and CD, BA and CD, AB and DC, BA and DC.
Thus, we multiply by 4:
4 * 3! = 24.

What about those arrangements where CD/DC are seated before AB/BA? You have 4 possibilities there as well, bringing the total to 8.

edge wrote:

GMATGuruNY wrote:

edge wrote:

GMATGuruNY wrote:...Total arrangements with AB and CD together means we're arranging the 3 elements AB, CD, and E = 3! = 6. Now we multiply by 4 because AB can be reversed, CD can be reversed, and both AB and CD can be reversed: 4*6 = 24...

Shouldn't we multiply by 8 because of 3 reversals?

The number of ways to arrange the 3 elements AB, CD and E = 3! = 6.
There are 4 variations in which AB and CD sit as couples: AB and CD, BA and CD, AB and DC, BA and DC.
Thus, we multiply by 4:
4 * 3! = 24.

What about those arrangements where CD/DC are seated before AB/BA? You have 4 possibilities there as well, bringing the total to 8.

I don't quite understand what you're saying.

Here are the arrangements in which AB and CD sit as couples:

AB CD E
BA CD E
AB DC E
BA DC E

CD AB E
CD BA E
DC AB E
DC BA E

AB E CD
BA E CD
AB E DC
BA E DC

CD E AB
CD E BA
DC E AB
DC E BA

E AB CD
E BA CD
E AB DC
E BA DC

E CD AB
E CD BA
E DC AB
E DC BA

Total = 4 * 3! = 24.

just deleting my post

Brent@GMATPrepNow wrote:

Tryingmybest wrote:Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

a.1/5 b 1/4 c 3/8 d 2/5 e 1/2

OA -[spoiler]D bit I get 2/15[/spoiler]

# of ways to sit the couple together: Glue the couple together. They are now one entity. Now remove a chair since this one entity still represents two people and, thus two chairs.
Seat the couple: 4 chairs --> 4 ways.
Seat the single person. 3 chairs remaining -->3 ways
The total number of ways to seat everyone is 4x3.
BUT there are two ways we can glue the couple together (MW or WM)
So, the total number of ways to seat everyone is 4x3x2=24

# of ways to sit everyone:
Seat one person. 5 chairs --> 5 ways
Seat next person. 4 chairs remaining --> 4 ways
Seat last person. 3 chairs --> 3 ways
Total = 5x4x3 = 60

Probability = 24/60 = 2/5

Plz Help me to fix my use of Anagram to solve the probem

Either of couple seat together(1)+ Both seat together (2) + Neither seat together = 1


To find Neither seat together, let find 1 & 2

* Find Either of couple seat together
- Probability to arrange all at random: Anagram ABCDE -->5! = 120
- Probability that one couple seat together, the other not:
anagram BCDE (pretend that couple AB is B) --> 4! = 24
there are 2 different persons in the couple --> 24x2 = 48
(But this may include 2 case: only one seat together, both couple incidently seat together)
--> 48/120

* Find Probability that both couples seat together
BCE (pretend that a couple is one word) --> 3! = 6
There are 4 different persons: 6x4 = 24
--> 24/120

48/120 + 24/120 + X = 1
72/120 + X = 1

X = 48/120 = 2/5

Can anyone help me to find the probability that Either of the couple seat together, since I thought my solution has flaw (I try to find 1 couple seat together, but this is not the only case since 2 couple may incidently seat together????)

I think here : anagram BCDE (pretend that couple AB is B) --> 4! = 24
there are 2 different persons in the couple --> 24x2 = 48
I should find: probability only one = Probability one couple seat together x probability the other not

GMATGuruNY wrote:

edge wrote:

GMATGuruNY wrote:

edge wrote:

GMATGuruNY wrote:...Total arrangements with AB and CD together means we're arranging the 3 elements AB, CD, and E = 3! = 6. Now we multiply by 4 because AB can be reversed, CD can be reversed, and both AB and CD can be reversed: 4*6 = 24...

Shouldn't we multiply by 8 because of 3 reversals?

The number of ways to arrange the 3 elements AB, CD and E = 3! = 6.
There are 4 variations in which AB and CD sit as couples: AB and CD, BA and CD, AB and DC, BA and DC.
Thus, we multiply by 4:
4 * 3! = 24.

What about those arrangements where CD/DC are seated before AB/BA? You have 4 possibilities there as well, bringing the total to 8.

I don't quite understand what you're saying.

Here are the arrangements in which AB and CD sit as couples...

Thanks; it seems I was double-counting. I understand now (hopefully correctly) that the possibility of the CD/DC couple switching with the AB/BA couple is accounted for in 3!.